A228229 Recurrence a(n) = n*(n + 1)*a(n-1) + 1 with a(0) = 1.
1, 3, 19, 229, 4581, 137431, 5772103, 323237769, 23273119369, 2094580743211, 230403881753211, 30413312391423853, 4744476733062121069, 863494765417306034559, 181333900737634267257391, 43520136177032224141773841, 11837477040152764966562484753
Offset: 0
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..252
- E. W. Weisstein, Modified Bessel Function of the First Kind
Programs
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Maple
A228229 :=proc(n) option remember if n = 0 then 1 else n*(n+1)*procname(n-1) + 1 end if: end proc: seq(A228229(n), n = 0..20);
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Mathematica
RecurrenceTable[{a[n] == n*(n + 1)*a[n-1] + 1, a[0] == 1},a,{n,0,20}] (* Vaclav Kotesovec, May 06 2015 *)
Formula
a(n) = n!*(n + 1)!*sum {k = 0..n} 1/(k!*(k + 1)!).
Generating function: 1/(1 - x)*1/sqrt(x)*BesselI(1, 2*sqrt(x)) = sum {n >= 0} a(n)*x^n/(n!*(n + 1)!).
Defining recurrence equation: a(n) = n*(n + 1)*a(n-1) + 1 with a(0) = 1.
Alternative recurrence equation: a(0) = 1, a(1) = 3, and for n >= 2, a(n) = (n*(n + 1) + 1)*a(n-1) - n*(n - 1)*a(n-2).
The sequence b(n) := n!*(n + 1)! satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 2. It follows that we have the finite continued fraction expansion a(n) = n!*(n + 1)!*(1/(1 - 1/(3 - 2/(7 - 6/(13 - … - n*(n - 1)/(n^2 + n + 1)))))). Taking the limit yields the continued fraction expansion for the modified Bessel function value BesselI(1,2) = sum {k = 0..inf} 1/(k!*(k + 1)!) = 1/(1 - 1/(3 - 2/(7 - 6/(13 - ...- n*(n - 1)/(n^2 + n + 1 - ...))))) = 1.590636... (see A096789).
a(n) ~ BesselI(1,2) * n!*(n+1)!. - Vaclav Kotesovec, May 06 2015
Comments